In this answer, it is stated that applying post-selection to quantum teleportation results in Alice communicating to Bob backwards in time. Could someone explain how this works?

I am particularly confused by how Alice decides what to post-select on. For example, if she wanted to teleport $$\newcommand{\ket}[1]{\lvert #1\rangle}\ket{-} = \frac{1}{\sqrt{2}}(\ket{0}-\ket{1})$$ then she could not post select on $\ket{00}$ because the amplitude of this state in the first two qubits would be zero.

So, it seems like what state she post-selects on depends on what state she wants to teleport. This means she needs to tell Bob what state she is going to post-select on after deciding what state to send him. This seems equivalent to sending him the two classical bits in the usual quantum teleportation.

What am I missing here?

  • $\begingroup$ I'm not sure this was your point, but to "post-select on a state" doesn't mean you decide what state is going to come out of the apparatus. You decide what measurement basis you want to use, and then you measure one of the possible outcomes. To "post-select" on some outcome means to wait until you get that outcome, and then only consider what happens in those cases. This is way a post-selection process is always probabilistic, and if the probability of measuring a state is zero, the corresponding post-selection occurs with zero probability $\endgroup$
    – glS
    Commented Aug 10, 2020 at 19:55
  • $\begingroup$ @glS in the linked answer, there is a distinction made between experimental and theoretical post-selection. I think you are referring to the experimental case, whereas in the theoretical case we assume there is an actual “post-selection operator” $\endgroup$
    – Owen
    Commented Aug 10, 2020 at 20:06
  • $\begingroup$ I don't think there is a real distinction, it's just a matter of what aspects of it are considered important in different communities. The associated complexity class is about what is possible when you don't take into account the "cost" of waiting for the post-selected state to come out, or in other words, you consider a "magical machine" that just outputs the result of the post-selection. It's the same idea, you are just considering different aspects of it. Regardless, in the context of teleportation that's certainly not the aspect you are interested in $\endgroup$
    – glS
    Commented Aug 10, 2020 at 20:14

1 Answer 1


As I understand it, the key caveat to postselection, at least in regards to retrocausal effects, is that Bob already knows what state Alice will post-select. Since he already knows what she will do, and she always does so with 100% success rate, then he can just treat his qubit as already having underwent the measurement and received a pure state, so that he can progress with any operations he intended to do once he had received the information immediately.

The "time travel" aspect comes into it because Bob doesn't actually need to wait for Alice to measure, so he can be said to have already received her measurement result before she made it, as she will always succeed in making it.


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